Commuting Invertible Matrices

Hey all,
I'm having trouble showing the following point:
if A is an invertible (n×n, real) matrix that commutes with ALL other invertible (n×n, real) matrices, then A is of the form cI, where c is any real number not equal to 0.

Thanks for the suggestion. So I had to use matrices B of the form I with an additional 1 at one other entry. By brute force, calculating the corresponding entries of AB would eventually give that all diagonal entries of A were the same, and that all other non diagonal entries must be zero. Sweet.

Here's a conceptual suggestion. Think of A as a transformation and let v be any non zero vector. We want to show first that Av = cv for some constant c.

Choose a basis involving v as first vector and define another transformation B that takes v to v and the other basis vectors to 0. Then applying AB = BA to v shows that ABv = Av = BAv, so B acts on Av != 0 as the identity. Since A is invertible, Av is not zero, so then Av = cv for some non zero constant c, since multiples of v are the only non zero vectors B takes to themselves.

Thus every vector v is an "eigenvector" for A, i.e. Av always equals cv for some c possibly depending on v. Now if every vector is an eigen - vector, we claim all the eigenvalues must be equal.

To see that, assume that Av = cv and Aw = dw, where v and w are in different directions, and look at A(v+w) = cv + dw = e(v+w), and check that we must have c=d=e.

Now why is that? well then cv + dw -ev -ew = 0 = (c-e)v + (d-e)w, hence
(c-e)v = (e-d)w, so these multiples of v and w are equal.

But since v, w are in different directions, their only equal multiples are zero, so d=e, c=e. QED.